Problem 29 - Thomas Calculus

Question

Suppose that \(f^{\prime}(x)=2 x\) for all \(x .\) Find \(f(2)\) if \begin{equation}\quad \text { a. }f(0)=0 \quad \text { b. } f(1)=0 \quad \text { c. } f(-2)=3.\end{equation}

Step-by-Step Solution

Answer

f(2) = 4 for (a), f(2) = 3 for (b) and (c).

Step 1: Understand the Derivative and Given Conditions

We are given the derivative of the function, \( f'(x) = 2x \). This implies that the original function \( f(x) \) is the antiderivative of \( 2x \). We need to find \( f(2) \) under different initial conditions: \( f(0) = 0 \), \( f(1) = 0 \), and \( f(-2) = 3 \).

Step 2: Find the General Form of the Function

The function \( f(x) \) is an antiderivative of \( 2x \). To find it, we integrate \( 2x \) with respect to \( x \): \[ f(x) = \int 2x \, dx = x^2 + C \]where \( C \) is the constant of integration.

Step 3: Apply Initial Condition \(f(0)=0\)

Substitute \( x=0 \) and \( f(x) = 0 \) into the expression \( f(x) = x^2 + C \).\[ 0 = 0^2 + C \]Solving for \( C \), we find:\[ C = 0 \]So, \( f(x) = x^2 \), and therefore, \( f(2) = 2^2 = 4 \).

Step 4: Apply Initial Condition \(f(1)=0\)

Substitute \( x=1 \) and \( f(x) = 0 \) into the expression \( f(x) = x^2 + C \).\[ 0 = 1^2 + C \]Solving for \( C \), we find:\[ C = -1 \]So, \( f(x) = x^2 - 1 \), and therefore, \( f(2) = 2^2 - 1 = 3 \).

Step 5: Apply Initial Condition \(f(-2)=3\)

Substitute \( x=-2 \) and \( f(x) = 3 \) into the expression \( f(x) = x^2 + C \).\[ 3 = (-2)^2 + C \]Solving for \( C \), we find:\[ C = -1 \]So, \( f(x) = x^2 - 1 \), and therefore, \( f(2) = 2^2 - 1 = 3 \).

Key Concepts

Antiderivative

When you are given a derivative, like in this problem where we have \( f'(x) = 2x \), finding the original function (or antiderivative) is your main goal. The process of obtaining an antiderivative is the inverse of taking a derivative. It's like retracing your steps to find the function that originally gave the derivative. In this example, we start with differentiating \( f(x) \) to get \( 2x \). To find the antiderivative, we integrate \( 2x \) with respect to \( x \), leading to the function \( f(x) = x^2 + C \). This C represents the constant of integration which can vary depending on initial conditions.

Constant of Integration

When we integrate a function, we always add a constant of integration, denoted by \( C \). This is crucial because integration is essentially an endless process of finding functions, and the derivative of a constant is zero. So, during integration, there could be any constant added to the function that still results in the same derivative. For example, when we find the antiderivative of \( 2x \), it results in \( f(x) = x^2 + C \). Without the initial or specific conditions, there are infinitely many antiderivatives, each differing by a constant. That's why we need initial conditions to pinpoint the exact function.

Initial Conditions

Initial conditions are specific values given for the function at certain points which help us determine the constant of integration, \( C \). They ensure that we pick the right specific function out of the infinite possibilities given by the antiderivative. In our exercise, we have three initial conditions:

• \( f(0) = 0 \)
• \( f(1) = 0 \)
• \( f(-2) = 3 \)

These conditions tell us what the function equals at these specific points, allowing us to solve for \( C \) in every scenario, and thus, we find the specific function for each case.

Integration

Integration is the process to find the antiderivative or the function whose derivative is given. It's the reverse operation of differentiation, and in essence, it is the process of adding up infinitely small quantities to retrieve a whole. In the example at hand, we integrate the derivative \( 2x \) to get the function \( f(x) = x^2 + C \). Integration gives us a family of functions each differing by a constant, which are then narrowed down through initial conditions to find the exact answer.

Differentiation

Differentiation is the process used to find the rate at which things change. It is used to determine the derivative of a function. The derivative is a measure of how a function changes as its input changes—it tells you the slope or steepness of the function at any given point.In this problem, we start by knowing the derivative \( f'(x) = 2x \). Differentiation applied to the original function \( f(x) \) produced this result. Understanding differentiation helps in the reverse process (integration) because it's about finding what the function looked like before its rate of change was calculated. In this case, going from \( f'(x) \) to \( f(x) \).